Intro to 2D Motion

Concept: Motion in 2D

21m
Video Transcript

hey guys we are now gonna talk about motion two dimension or motion in a 2-dimensional plane and I'm gonna show how itÕs very similar to motion in 1 dimension, we can use lot of the same equations but it is a little more complicated so letÕs check it out. Alright so motion in 2 dimensions or motion in a plane is very similar to one dimensional motion, now one dimensional motion can also be thought of motion in a straight line so it is when we are moving left and right in the x-axis or up and down the y-axis we are restricted that you can only move this way or this way . Now we are gonna moving on a plane so imagine this as a table top so you have your x y plane and you are moving at an angle like this okay . We are also gonna be moving in curved paths and things like that so we can think of 2-dimensional motion as two sets of 1 dimension. So one problem of 2 dimension is really two problems of 1 dimension, we are gonna break it down okay. So for example here I have my y-axis and then I have my x-axis and if I am going from a to b I can break this down into motion along the x only like that plus motion along the y only okay and I can call this delta x because my change in position in the x-axis delta y change in position in y-axis and I am gonna call this my delta r which is a 2 dimensional displacement okay. These motions here in the x and the y they are independent of each other so whatever happens in the x-axis has no relation to what happens on the y-axis except that they are synchronized which means the time elapse is the same for both same time okay so they are happening at the same time but they are independent from each other , what happens in the x-axis doesnÕt affect the y-axis and vice-versa. In physics whenever anything happens in 2 dimensions will actually go into first break it down back to 1 dimension. Everything that we solve in physics will be solved in 1 dimension, so when we have a 2 dimensional problem we decompose into x and y components much like what we did with vectors. If you have vector then it will break it into x and y okay so position, displacement, velocity and acceleration are all vectors and they may have to be decompose if they are at an angle so letÕs check out how it works for each one of them. Position another definition of position, it is a vector from the origin, origin is over here where you have a 0 comma 0. ItÕs a vector from the origin to aor so if you are at point p your position vector looks like this okay and because its at an angle I can break it down into one vector in the x axis and one vector in the y axis I'm gonna call this guy here delta x I am sorry its actually position so I'm gonna call this guy just x which is the position vector in the x axis this arrow this is y and this is r okay because this is position so position in the x is just x arrow y arrow and the 2 dimensional vector is r arrow. There is an angle here that would make with the x-axis , we always make our angles with the x axis remember and there are four equations that relate these four variables here r, x, y and theta and these are the vector composition decomposition equations. So they look like this. The magnitude of the r vector how long this red line is , is given by the square root of x square plus y square . This angle that I am showing here is given by if you remember the r tangent of y over x. The other two equations that acting at x from r and theta , x equals rcosine of theta and y equals rsine of theta and these are just our vector equations. Okay, displacement is not from origin to point but between two points, itÕs the shortest line , the shortest path between two points so you just draw straight line between them , this is in 2 dimension so you can break this into x only and y only or the combination of both, this is delta x and this is delta y this is delta r and I have a theta here so notice that here its x, y, r and here it is delta x, delta y, delta r because itÕs the change in position right, change in position and the equation is very similar, the magnitude of delta r is just the square root of delta x square delta y square , so basically itÕs the same thing just put deltas everywhere. Okay and then to find delta x obviously, delta x is just delta rcosine of theta and delta y is just delta rsine of theta nothing shocking there. What happens with the velocity vector. So letÕs say I am moving this way if I am moving in this direction here then that is my velocity vector . I can break those down because it is going at an angle and I am gonna do that so hereÕs the theta here and I am gonna break this down into its Vx and its Vy. Okay so this is Vx and this is Vy. Now V is the change in position over change in time so itÕs the dispalcement over time but if itÕs a 2-dimensional V itÕs not delta x itÕs not delta y its delta r right so delta r over delta t is which you gonna get on a 2-dimensional velocity okay. Another way that you can find velocity is if you know your Vx and your Vy, so you just do this, Pythagorean theorem right. So how do I get Vx theres two way you can get Vx. Well if the two dimensional V was the 2 dimensional displacement divided by time , then the x velocity is the x displacement divided by time , just like how your y velocity is your y displacement divided by time. Okay the other way you can do this is with vector decomposition so Vx is Vcosine of theta and you should imagine that Vy would be Vsine of theta, the same thing happens with acceleration where I can break this down into Ax and Ay okay and Ae is the change in velocity, in this case V is a 2 dimensional velocity not Vx not Vy just V over the cahnge in time and aging this can be the square root of Ax square Ay square so combination of its components Ax how do I get Ax , how do I get Ay . Ax and Ay they follow very similar to Vx and Vy so instead of the change in velocity is the change velocity in the x-axis divided by time and this is the change in velocity in the y-axis divided by time. Another way you can do this is Acosine of thta Asine of theta. I know I gave you a lots of equations but I hope you realise that lot of these are just vector equations, so whats actually new here, this is kind of new and the idea that you do yous Vx and Vy this way okay , other than that the only other new ones are these two which are basically the same thing that we saw in the previous ones , this is just the definition of acceleration. So theres one more thing that two more things I wanna show you so in 2 dimensions your velocity is V, your velocity in the x-axis is Vx , this is Vy same thing here Ax , Ay and if going on a circle which is a special case of 2 dimensional motion , if you are going on a circle , every point around a circle you are gonna have a velocity this way , we call it tangentional velocity because it is tangent to the path so for example here this velocity would look like this okay so its always tangent to the path here this velocity will look like this just imagine that you have a objevct spinning on a rope and the rope breaks right here the object will be going that way right so thats your tangential velocity and at every point you also have a centripetal acceleration , centripetal acceleration I am gonna call that Ac we will talk about this okay. So and there an equation that relates those two , we will leave it for later. Alright so lets work out this example , it says here that the diagram shows your position and meters as you move in x-y plane so obviously this is your Y-axis and this is your X-axis so imagine you are on a surface and you are moving right at different angles. So it says that it takes 10 seconds to go from A to B so let me right this down somewhere here the time betweeen A and B is 10 seconds , this is the first interval and it takes 5 seconds to go from B to C while moving with a different constant velocity , so its basically staying at a constant velocity here and a constant velocity here but that is different okay . So the firstly it says its drop position vector is thats your x, y, r right here position vectors x, y ,r for point A and then displacement vectors for AB and BC. Lets do position vector first. Remember position vector is a vector that goes from the origin to where you are, so its just an arrow from here to here , now I want the r vector which is this but I also want the x and y vector in other words the 1 dimensional components of the r vector . So lets do that real quick , this would be my x over here and this would be my y over here. Okay and I can figure how big how long they are by just looking at their grid. So this is in meters it says so this is notice how this is a force , so this is a two over here this is a six over here so every grid isa 2 by 2 grid so the magnitude of x is 6 and the magnitude of y is 4 and they are both positive alright . So I can figure out the total lenth of the r vector and thats what part b says , part b is gonna ask me to calculate the magnitude and direction of all these different vectors okay , so let me find the magnitude of r this is just if you notice it forms a triangle , this is just pythagorean theoram , so I get x square plus y square so it is the square root of 6 square plus 4 square and if you plus from the calculator you get 7.21 meters everything is in meters so that what you get , if you want the angle for position A so you can call this Ra if you want and theta a then you gonna do what the arc tangents so by now you should know this pretty well , y over x and if you do that you get the arc tangents of 4 over 6 and it is 33.7 degrees, this is in the first quadrant right , this angle is in the first quadrant if you think it as a unit circle so it occurs in the first quadrant I dont have to do anything with that angle . So now I wanna right , I wanna draw displacement vectors so not position right position is origin to point , displacement is A to B so they are gonna be delta x , delta y and delta r so x-axis, y-axis in 2 dimensional. For intervals AB and BC so we get a different color here , from A to B obviously this guy here is my delta r from A to B and if I want to draw my delta x and my delta y here is my delta x from A to B , I'm not gonna draw my delta y here , I'm gonna draw my delta y over here now it makes no difference their is a reason why I am doing it there and its just I have to draw everything together eventually , itÕs gonna be a little messy . So whats delta x , theres two squares here so this is a 4 and its to the right so itrs positive , delta y is going through four squares so this is an eight alright and once again if I wanna find the magnitude , the length of this blue line here theres goes an angle and the magnitude of delta r from A to B is just the square root of the sum of the squares of its components so just the 4 and the 8 here , Pythagorean theorem with the 4 and the 8 and if you plug this into the calculator you get 8.94 meters and if I am looking for the angle betwwen A and B which is this guy over here then I can just draw the arc tangents of y over x look closely y is 8 this is a 4 so its the arc tangent of 8 over 4 wich is 63.4 degrees okay, so we are done with A to B thats all the information now B to C is this is obviously my delta r right here from b to c and this is my delta x remember its the one way to think about is the x component is the shadow of that so delta x from B to C looks like this and if I count its 6 and delta y is this delta y BC and if you count that is a 4 okay so the magnitude of delta r on my 2 dimensional displacement from B to C is again given by the square root of its components I got a 6 square plus 4 square its the same numbers we had here if you notice , exact same numbers so I am just gonna get 7.21 in the angle from B to C is again the arc tangent of y over x and in this case the arc tangent of 4 over 6 once again so it is 33.7 degrees once again.

These angles are all in the first quadrant so far so I donÕt have to do anything with them and then the last piece of the problem is asking you, that it says calculate the magnitude and direction of displacement velocity vectors for intervals AB, BC and AC so wanna know how longer these vectors , the displacement vectors as well as the velocity vectors , so letÕs finish this . So I have already done displacement for A and B so I just have to do displacement for A to C okay so A to C is gonna look like this let me do a different color heres A to C the green line right so this is delta r for AC , delta x for AC looks like this so I told you that it will get a little ugly and so this is delta x for AC because its the displacement in the x- axis alone and this is delta y and its cut through here , this is delta y AC and I can just kind of eyeball this and measure right , we have done this , this is basically vector addition so have one, two, three, four, five so this is ten and then on y-axis it goes all the way from down here so I have one, two, three, four, five, six and this is twelve and they are both positive okay so delta r from A to C is the square root of ten square plus twelve square and if you plug this in you get 15.6 meters and for the angle again its just the arc tangent of y over x so y is twelve, x is ten and if you do that you get a 50.2 again first quadrant I dont have to touch it and we are done with finding the displacement vector , we are almost done .

Now we do have to find the velocity vectors , so this is obviously a very long question but itÕs just we get to see how this stuff actually works in practice okay. So now I wanna find the velocity between A and B , the velocity between B and C first half , second half and then the velocity for the entire interval. So how can I do that, if you look at your equations one of the ones you can use is that the velocity from and in 2 dimensions itÕs just the 2 dimensional displacement which is delta r over delta t . So I can just say the velocity between A and B the average velocity is AB delta t AB same thing here delta r from BC , delta t from BC and the same thing here delta r from AC to delta t AC okay , the biggest thing here is using the right numbers for the right the numbers that belong to the interform right so delta r for AB is this guy right here so its 8.94 and the time that it takes is 10 seconds so the answer here is that this is .894 meters per second okay . Here delta r PC is 7.21 the time is 5 seconds it says right up here 5 seconds , they give us that time and if you do this you get a 1.44 meters per second from A to C its not the average of these two its not the addition of these two numbers could rather remember you have to work out the entire distance the entire diaplacement rather which is 15.6 and the entire time so though you dont add the distances , you dont add the displacements here because its vector addition you have to figure how long this green line is okay you do add the times right because its just 10 plus 5 seconds so let me put it here like this 10 plus 5 seconds okay and if you combine that you get that this is 1.04 meters per second alright so thats it for this one , hopefully this helps a little bit straight , a little bit better how motion to 2 dimension works and thats it for this one .

Problem: An object moves up with a constant 8 m/s while moving to the left with a constant 6 m/s. Calculate the magnitude and direction of the object’s (total) velocity.

3m

Problem: An object is launched with an initial velocity of 50 m/s directed 37 degrees above the horizontal. Calculate the X and Y components of the object’s initial velocity.

1m

Problem: A car is initially at rest at (0, 0) meters on an X-Y plane. The car then accelerates for 4 s with a constant 6 m/s2 directed in the +x-axis. The car then accelerates for 5 s with a constant 5 m/s2 directed at 53° above the +x-axis. Find the magnitude and direction of the car’s velocity after the two accelerations (after 8 s → hint: use v = vo + at).

9m

Intro to 2D Motion Additional Practice Problems

A hockey puck slides horizontally off the edge of a table with an initial velocity of 20.0 m/s. The height of the table above the ground is 2.00 m. What is the magnitude of the vertical component of the velocity of the puck just before it hits the ground?

[A] 6.26 m/s

[B] 12.5 m/s 

[C] 16.3 m/s

[D] 19.6 m/s 

[E] 22.4 m/s

Watch Solution

A motorcycle rider wants to try the following stunt: she will start from rest and accelerate her bike at constant acceleration a0 on a horizontal platform of length L; she will then jump with her bike trying to land on a second platform placed a distance y0 below the first one.

[a] What is the speed of the rider at the end of the first platform?

[b] What is the maximum distance at which the second platform can be placed for the rider to reach it?

[c] What is the speed of the rider when she reaches the second platform?

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The velocity of an airplane as a function of time can be written as  v(t) = b− 3ct2j where b and c are positive constants. Provide your answers in terms of b, c, and t when necessary.

[a] What are the SI units of b and c?

[b] Find an expression for the position r(t) of the airplane as a function of time assuming that the airplane is at the origin at t = 0.

[c] Find an expression for the acceleration a(t) of the airplane as a function of time

[d] Find the trajectory of the airplane (y vs. x).

 

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Using a reference frame with the origin at the take-off airport, the positive x-axis due East, and the positive y-axis due North, the acceleration a of an airplane as a function of time can be described as:

a = (αt)î + (βt 4 − γt)ĵ,

with α, β, and γ positive and constant.

Assuming that the airplane takes off from the airport at time t = 0 with zero initial velocity:

a) What are the units of α, β, and γ?

b) Find the time(s) when the airplane position is directly NE of the airport.

c) Find the trajectory of the plane, y(x).

Write your results in terms of α, β, and γ. Remember to check the dimensions/units for each answer.

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The velocity vector v(t) measured in m/s for a particle at time t is given as  v(t) = [(3t + 4t3)î − 2ĵ] m/s, where t is in seconds.

[a] Find the units of the coefficients 3, 4, and 2 in the vector  v.

[b] Find the average acceleration in the interval t = 0 to t = 2 seconds.

[c] Find the acceleration at t = 2 s.

[d] Say that at t = 0 the particle is at the origin, r(0) = 0. Find the position vector  r(t).

[e] Eliminate t to give the equation of the trajectory x as a function of y.

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A bullet is fired horizontally, and at the same instant a second bullet is dropped from the same height. Ignore air resistance. Compare the times of the fall of the two bullets.

A) They hit at the same time.

B) The fired bullet hits first.

C) The dropped bullet hits first.

D) Cannot tell without knowing the masses.

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A passenger on a Ferris wheel moves in a vertical circle of radius  R with constant speed v. Assuming that the seat remains upright during the motion, derive an expression for the magnitude of the upward force the seat exerts on the passenger at the bottom of the circle if the passenger's mass is m.

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When fighting forest fires, airplanes work in support of ground crews by dropping water on the fires. A pilot is practicing by dropping water on the fires. A pilot is practicing by dropping a canister of red dye, hoping to hit a target on the ground below. If the plane is flying in a horizontal path of 41.0 m above the ground and with a speed of 18.0 m/s, at what horizontal distance from the target should the pilot release the canister? Ignore air resistance.

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Two balls roll off a 0.8 m high table with horizontal velocities of 2.0 m/s and 4.0 m/s, respectively. What is the horizontal velocity of the faster ball when it hits the ground?

A) 9.80 m/s

B) 3.96 m/s

C) 0.80 m/s

D) 2.00 m/s

E) 4.00 m/s

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Two balls roll off a 0.8 m high table with horizontal velocities of 2.0 m/s and 4.0 m/s, respectively. They reach the edge of the table at the same time. Which ball hits the ground first?

A)  Both balls hit the ground at the same time.

B) The slower ball hits the ground first.

C) The ball that travels the shorter distance hits the ground first.

D) The faster ball hits the ground first.

Watch Solution

Superman's arch nemesis Lex Luthor pushes Lois Lane horizontally with a speed of 5 m/s off a 35 m tall building. Superman stands on the ground just below the edge of the rood. Ignore air resistance.

What will be the vertical component of Lois' velocity just before Superman catches her?

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Superman's arch nemesis Lex Luthor pushes Lois Lane horizontally with a speed of 5 m/s off a 35 m tall building. Superman stands on the ground just below the edge of the rood. Ignore air resistance.

How far from the building will he catch her?

Watch Solution

Superman's arch nemesis Lex Luthor pushes Lois Lane horizontally with a speed of 5 m/s off a 35 m tall building. Superman stands on the ground just below the edge of the rood. Ignore air resistance.

How fast must he run (at constant speed) to catch Lois just before she hits the ground?

Watch Solution

When fighting forest fires, airplanes work in support of ground crews by dropping water on the fires. A pilot is practicing by dropping a canister of red dye, hoping to hit a target on the ground below. If the plane is flying in a horizontal path 29.0 m above the ground and with a speed of 19.0 m/s, at what horizontal distance from the target should the pilot release the canister? Ignore air resistance.

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Two balls roll off a 1.1 m high table with horizontal velocities of 2.5 m/s and 5.0 m/s, respectively. they reach the edge of the table at the same time. What is the horizontal velocity of the faster ball when it hits the ground? 

A) 5.00 m/s

B) 1.10 m/s

C) 2.50 m/s

D) 4.64 m/s

E) 9.80 m/s

Watch Solution

Two balls roll off a 1.1 m high table with horizontal velocities of 2.5 m/s and 5.0 m/s, respectively. they reach the edge of the table at the same time. Which ball hits the ground first?

A) The ball that travels the shorter distance hits the ground first.

B) The slower ball hits the ground first.

C) The faster ball hits the ground first.

D) Both balls hit the ground at the same time.

Watch Solution

A stuntman for a 007 movie scene needs to jump into the lake below while avoiding the ledge.

1) If his initial velocity is horizontal, what is the time for the stuntman to fall 9.0m?

2) What must be the minimum speed vo to pull off such a stunt?

3) The stuntman knew that he can only run as fast as 1.2 m/s, and decided to perform this stunt by jumping with an upward angle of 45 degrees. What will be his landing point, and will that be greater than 1.75 meters?

Watch Solution